Let's stay that Steph makes 90% of his free throws while LeBron makes 70%. If they each shoot 100 free throws, what is the probability that Steph makes more free throws than LeBron?
I think it's fair to assume that the number of free throws Steph makes is a binomial random variable $X \sim Binomial(100, 0.9)$. Similarly, the number LeBron makes is $Y \sim Binomial(100, 0.7)$. So, I am trying to find out $P(X > Y)$. If it helps, we could define the difference as $D = X - Y$ and then I am trying to find $P(D > 0)$. This is where I get stuck - how do I combine two binomial distributions to figure out how $D$ is distributed?
Also, would the question be more difficult (or different) to answer if the two players shot different numbers of free throws?
Say, $X$ is the random variable for Steph's making his free throws and $Y$ for LeBron.
If you are looking for the exact form, it will look as below -
$\displaystyle P(X \gt Y) = \sum \limits_{k=1}^{n} {n \choose k} \, (0.9)^k \, (0.1)^{(n-k)} \, \sum \limits_{i=0}^{k-1} {n \choose i} (0.7)^i \, (0.3)^{(n-i)}$
For $X = k$ successes, we know that $0 \leq Y \leq k-1 \,$ are acceptable.
Distribution for just $n = 2$ is
$P(X \gt Y) = 0.4293, P(X \lt Y) = 0.0973, P(X = Y) = 0.4734$.
Checking WolframAlpha for $n = 10$ throws, I get $P(X \gt Y)$ as $\approx 0.81$ which was $0.4923$ for $n = 2$.